A certain metallic surface is illuminated with monochromatic light of wavelength, $\lambda.$ The stopping potential for photoelectric current for this light is $3V_0.$ If the same surface is illuminated with light of wavelength $2\lambda,$ the stopping potential is $V_0.$ The threshold wavelength for this surface for photoelectric effect is

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(c =$velocity of light$)$

An electron of mass m and a photon have the same energy E. The ratio of de Broglie wavelengths associated with them is

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is (Given $h = 4.14 \times 10^{-15}$ eVs and $c = 3 \times 10^8$ m/s)

The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m is

An electron of mass $m$ with an initial velocity $\mathbf{v} = v_0 \hat{i}$ (where $v_0 > 0)$ enters an electric field $\mathbf{E} = -E_0 \hat{i}$ (where $E_0$ is constant and greater than zero) at $t = 0.$ If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time $t$ is

When the light of frequency $2\nu_0$ (where $\nu_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1.$ When the frequency of the incident radiation is increased to $5\nu_0,$ the maximum velocity of electrons emitted from the same plate is $v_2.$ The ratio of $v_1$ to $v_2$ is

An electron is accelerated through a potential difference of $10,000$ V. Its de Broglie wavelength is, (nearly):$\quad (m_e = 9 \times 10^{-31}$ kg$)$

Light of frequency $1.5$ times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}$ nm, the potential difference is:

The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600$ nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be:$\newline (h=6.6 \times 10^{-34}$ Js$)$

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If $m$ is the mass of the photoelectron emitted from the surface, the de-Broglie wavelength $\lambda_d$ of the electron is given by:

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is?

When two monochromatic light of frequency $\nu$ and $\frac{\nu}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_s}{2}$ and $V_s$ respectively. The threshold frequency for this metal is:

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:

The work function of Caesium (Cs), Potassium (K), and Sodium (Na) are 2.14 eV, 2.30 eV, and 2.75 eV respectively. If incident electromagnetic radiation has an energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?

If $c$ is the velocity of light in free space, the correct statements about photon among the following are:$\newline$A. The energy of a photon is $E = h\nu. \newline$B. The velocity of a photon is $c. \newline$C. The momentum of a photon $p = \frac{h\nu}{c} \newline$D. In a photon-electron collision, both total energy and total momentum are conserved.$\newline$E. Photon possesses positive charge.

The graph which shows the variation of $\frac{1}{\lambda^2}$ and its kinetic energy, $E$ (where $\lambda$ is de Broglie wavelength of a free particle) :